Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781305176461
Author: Kotz
Publisher: Cengage
Question
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Chapter 21.2, Problem 1CYU

(a)

Interpretation Introduction

Interpretation: To write the balanced chemical equation for the formation of NaBr.

Concept Introduction: Main group elements are categorised as s-block and p-block elements.  The s-block elements includes metals belonging to group 1 and group 2 and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. The metal gets positively charge and the non metal attains a negative charge thus forming cations and anions respectively. They do so to attain a noble gas configuration or to attain stability.

  MMn++ne

Here, by losing electrons metal M achieve the noble gas configuration. These electrons are gained by the non-metals X as shown below.

  X+neXn

The metals of group IA form +1 ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group IA elements is +1.  Similarly, group IIA elements form +2 ions by losing two electrons and have an oxidation number of +2. The non-metal gains these electrons to form anions with 1 and 2 charge.

The stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation  if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of CaSe should be written. 

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group 1 and group 2 and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. The metal gets positively charge and the non metal attains a negative charge thus forming cations and anions respectively. They do so to attain a noble gas configuration or to attain stability.

  MMn++ne

Here, by losing electrons metal M achieve the noble gas configuration. These electrons are gained by the non-metals X as shown below.

  X+neXn

The metals of group 1A form +1 ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group IA elements is +1.  Similarly, group IIA elements form +2 ions by losing two electrons and have an oxidation number of +2. The non-metal gains these electrons to form anions with 1 and 2 charge.

The stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(c)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of PbO should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group 1 and group 2 and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. The metal gets positively charge and the non metal attains a negative charge thus forming cations and anions respectively. They do so to attain a noble gas configuration or to attain stability.

  MMn++ne

Here, by losing electrons metal M achieve the noble gas configuration. These electrons are gained by the non-metals X as shown below.

  X+neXn

The metals of group IA form +1 ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group 1A elements is +1.  Similarly, group IIA elements form +2 ions by losing two electrons and have an oxidation number of +2. The non-metal gains these electrons to form anions with 1 and 2 charge.

The stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(d)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of AlCl3 should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group 1 and group 2 and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. The metal gets positively charge and the non metal attains a negative charge thus forming cations and anions respectively. They do so to attain a noble gas configuration or to attain stability.

  MMn++ne

Here, by losing electrons metal M achieve the noble gas configuration. These electrons are gained by the non-metals X as shown below.

  X+neXn

The metals of group IA form +1 ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group IA elements is +1. Similarly, group IIA elements form +2 ions by losing two electrons and have an oxidation number of +2. The non-metal gains these electrons to form anions with 1 and 2 charge.

The stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation  if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

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Chapter 21 Solutions

Chemistry & Chemical Reactivity

Ch. 21.8 - Prob. 4QCh. 21.8 - Prob. 3RCCh. 21.11 - Prob. 1QCh. 21.11 - Prob. 2QCh. 21 - Give examples of two basic oxides. Write equations...Ch. 21 - Prob. 2PSCh. 21 - Prob. 3PSCh. 21 - Prob. 4PSCh. 21 - Prob. 5PSCh. 21 - Prob. 6PSCh. 21 - For the product of the reaction you selected in...Ch. 21 - For the product of the reaction you selected in...Ch. 21 - Prob. 9PSCh. 21 - Prob. 10PSCh. 21 - Place the following oxides in order of increasing...Ch. 21 - Place the following oxides in order of increasing...Ch. 21 - Prob. 13PSCh. 21 - Prob. 14PSCh. 21 - Prob. 15PSCh. 21 - Prob. 16PSCh. 21 - Prob. 17PSCh. 21 - Prob. 18PSCh. 21 - Prob. 19PSCh. 21 - Prob. 20PSCh. 21 - Prob. 21PSCh. 21 - Write balanced equations for the reaction of...Ch. 21 - Prob. 23PSCh. 21 - (a) Write equations for the half-reactions that...Ch. 21 - When magnesium bums in air, it forms both an oxide...Ch. 21 - Prob. 26PSCh. 21 - Prob. 27PSCh. 21 - Prob. 28PSCh. 21 - Calcium oxide, CaO, is used to remove SO2 from...Ch. 21 - Prob. 30PSCh. 21 - Prob. 31PSCh. 21 - The boron trihalides (except BF3) hydrolyze...Ch. 21 - When boron hydrides burn in air, the reactions are...Ch. 21 - Prob. 34PSCh. 21 - Write balanced equations for the reactions of...Ch. 21 - Prob. 36PSCh. 21 - Prob. 37PSCh. 21 - Alumina, Al2O3, is amphoteric. Among examples of...Ch. 21 - Prob. 39PSCh. 21 - Prob. 40PSCh. 21 - Describe the structure of pyroxenes (see page...Ch. 21 - Describe how ultrapure silicon can be produced...Ch. 21 - Prob. 43PSCh. 21 - Prob. 44PSCh. 21 - Prob. 45PSCh. 21 - Prob. 46PSCh. 21 - Prob. 47PSCh. 21 - The overall reaction involved in the industrial...Ch. 21 - Prob. 49PSCh. 21 - Prob. 50PSCh. 21 - Prob. 51PSCh. 21 - Prob. 52PSCh. 21 - Prob. 53PSCh. 21 - Prob. 54PSCh. 21 - Prob. 55PSCh. 21 - Sulfur forms a range of compounds with fluorine....Ch. 21 - The halogen oxides and oxoanions are good...Ch. 21 - Prob. 58PSCh. 21 - Bromine is obtained from brine wells. The process...Ch. 21 - Prob. 60PSCh. 21 - Prob. 61PSCh. 21 - Halogens combine with one another to produce...Ch. 21 - The standard enthalpy of formation of XeF4 is 218...Ch. 21 - Draw the Lewis electron dot structure for XeO3F2....Ch. 21 - Prob. 65PSCh. 21 - Prob. 66PSCh. 21 - Prob. 67GQCh. 21 - Prob. 68GQCh. 21 - Consider the chemistries of the elements...Ch. 21 - When BCl3 gas is passed through an electric...Ch. 21 - Prob. 71GQCh. 21 - Prob. 72GQCh. 21 - Prob. 73GQCh. 21 - Prob. 74GQCh. 21 - Prob. 75GQCh. 21 - Prob. 76GQCh. 21 - Prob. 77GQCh. 21 - Prob. 78GQCh. 21 - Prob. 79GQCh. 21 - Prob. 80GQCh. 21 - Prob. 81GQCh. 21 - Prob. 83GQCh. 21 - Prob. 84GQCh. 21 - A Boron and hydrogen form an extensive family of...Ch. 21 - In 1774, C. Scheele obtained a gas by reacting...Ch. 21 - What current must be used in a Downs cell...Ch. 21 - The chemistry of gallium: (a) Gallium hydroxide,...Ch. 21 - Prob. 89GQCh. 21 - Prob. 90GQCh. 21 - Prob. 91GQCh. 21 - Prob. 92GQCh. 21 - Prob. 93ILCh. 21 - Prob. 94ILCh. 21 - Prob. 95ILCh. 21 - Prob. 96ILCh. 21 - Prob. 97ILCh. 21 - Prob. 98ILCh. 21 - Prob. 99SCQCh. 21 - Prob. 100SCQCh. 21 - Prob. 101SCQCh. 21 - Prob. 102SCQCh. 21 - Prob. 103SCQCh. 21 - Prob. 104SCQCh. 21 - Prob. 105SCQCh. 21 - Prob. 106SCQCh. 21 - Prob. 107SCQCh. 21 - Prob. 108SCQCh. 21 - Prob. 109SCQCh. 21 - Prob. 110SCQCh. 21 - Comparing the chemistry of carbon and silicon. (a)...Ch. 21 - Prob. 112SCQCh. 21 - Xenon trioxide, XeO3, reacts with aqueous base to...
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